Fluorine-18 undergoes positron emission with a half-life of 110. minutes. If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.)单项选择题
A
174 min
B
300. min
C
99 min
D
211 min
E
132 min

Question

Fluorine-18 undergoes positron emission with a half-life of 110. minutes. If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.)单项选择题

A

174 min

B

300. min

C

99 min

D

211 min

E

132 min

BlackTom AI · Accepted Answer

Radioactive decay follows first-order kinetics with N(t) = N0 · (1/2)^(t / T1/2), where T1/2 is the half-life. Here T1/2 = 110.0 min, N0 = 248 mg, and we seek t when N(t) = 83 mg.

1. Relevant concepts
- Exponential decay governed by the half-life formula N(t) = N0 · (1/2)^(t/T1/2).
- Solving for time requires taking the natural logarithm of both sides.

2. Step-by-step reasoning and algebra
Start with the decay equation:
83 = 248 · (1/2)^(t / 110)

Divide both sides by 248:
83 / 248 = (1/2)^(t / 110)

Compute the ratio:
83 / 248 = 0.334677419...

Take natural logs:
ln(0.334677419) = (t / 110) · ln(1/2)

Solve for t:
t = 110 · ln(0.334677419) / ln(1/2)

3. Intermediate calculations
ln(0.334677419) ≈ -1.094589
ln(1/2) = -0.693147
t = 110 · (-1.094589) / (-0.693147) ≈ 110 · 1.579422 ≈ 173.7364 min

Rounding gives t ≈ 174 min.

4. Why this answer is correct
The calculation uses the exact half-life exponential relationship and logarithms to invert the exponential. The numeric steps show the patient’s fluorine-18 falls from 248 mg to 83 mg in about 173.7 minutes, which rounds to 174 min.

5. Why the other options are incorrect
- 99 min: shorter than one half-life (110 min); after 99 min the amount would still exceed one half-life remaining (~>124 mg), so 99 min is too short.
- 132 min: between 110 and 174 but too small — direct calculation shows 132 min leaves more than 83 mg.
- 211 min and 300 min: both lie beyond the needed time; 211 min would give less than 83 mg (closer to two half-lives), and 300 min is well past that.

Conclusion: The computed time is approximately 174 min, matching the provided correct answer (174 min).

Fluorine-18 undergoes positron emission with a half-life of 110. minutes. If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.)单项选择题

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